题目:
Implement the following operations of a stack using queues.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- empty() -- Return whether the stack is empty.
Notes:
- You must use only standard operations of a queue -- which means only
push to back
,peek/pop from front
,size
, andis empty
operations are valid. - Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
- You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
提示:
此题要求用“队列”实现“栈”,我们可以通过将队列进行循环来实现栈的特性。比如在push()的时候,被push元素之前的所有元素都重新pop出来再push到队尾,经过这一操作之后,队列中的元素顺序就能和栈的“先进后出”的顺序一致了。
通过这种方法,push是O(n)的,而pop,top,empty都是O(1)的。
代码:
class Stack { queue<int> q; public: // Push element x onto stack. void push(int x) { q.push(x); for (int i = 0; i < q.size() - 1; ++i) { q.push(q.front()); q.pop(); } } // Removes the element on top of the stack. void pop() { q.pop(); } // Get the top element. int top() { return q.front(); } // Return whether the stack is empty. bool empty() { return q.empty(); } };